Suppose f is a differentiable function such that f′(x)≤2 for all x∈[1,3]. if f(1)=4, the mean value theorem says that f(3)≤v for what value of v? (choose v as small as possible.)

The mean value theorem is used to link the average rate of change and the derivative of a function.

The value of V is 8.

The given parameters are:

Mean value theorem states that:

If [a,b] and

(a,b),

Then there is a point c in (a,b), such that:  

From the question, we understand that: f is differentiable

This means that:

So, we have:

Substitute 4 for f(1)

Recall that:

The equation becomes

Cross multiply

Add 4 to both sides

From the question, we have:

By comparisons;

Hence, the value of V is 8.

Read more about mean value theorems at:

link

The MVT guarantees the existence of such that

Since for all , we have

so that .

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116